To find the charge distribution of proton , we study electron proton scattering and compute the form factor to find the cross section. The form factor comes out to be Fourier transform of charge distribution. It is then multiplied by the cross section calculated under assumption that proton is point particle. What I don't understand is why the form factor has to be Fourier transform of charge distribution. I don't want the derivation but some intuition.


1 Answer 1


You have surely seen slit experiments, where waves which pass through the slit and scatter to produce a pattern on a screen placed behind the slits. In the far-field approximation (also known as Fraunhofer diffraction) this pattern is precisely the Fourier transform of the slits. Perhaps you even remember that waves passing by lines produce the same pattern as waves passing through slits, given that the line and the slit have the same dimensions.

We can also play this game with holes and points or any kind of shape, really. The proton charge distribution is such a shape and the electron is a quantum mechanical wave passing along it. Like every other wave passing along an object, the wave will scatter and it is described by the Fourier transform of the object. The far field approximation always holds in a realistic experiment, because the the charge distribution is far smaller than the travel distance of the electrons.

  • $\begingroup$ Thank you for the answer. Can you explain why exactly we should expect that interference pattern should be fourier transform of slit in far-field approximation? $\endgroup$ Commented Aug 6, 2015 at 12:52
  • $\begingroup$ Essentially in the far-field you just need to sum up a wave originating from each point in the slit. If you describe each wave as an exponential $e^{-ikx}$ then you can add up these waves by integrating the $e^{-ikx}$ multiplied by the slit. This is exactly the Fourier transform of the slit, ignoring constants etc. $\endgroup$ Commented Aug 6, 2015 at 13:01
  • $\begingroup$ Chris is right. Electrons form waves with spherical wave fronts as they are released from the source. In the far-field approximation we assume that the wave fronts that hit the target are planar, not spherical (equivalently, we assume the "electron rays" are parallel). This is approximately true when the source and the target are separated by a large distance, compared to size of the target. Because the waves are planar we can simply add them up, as was mentioned by Chris. $\endgroup$
    – vosov
    Commented Aug 6, 2015 at 17:03

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